The construction employs a well known property of angle bisectors in a triangle. In ACD, let DP be the angle bisector of angle D. Then
(1)
AP/PC = AD/CD
It follows from (1) that if CD = N·AD, then AP = PC/N, or PC = N·AP, so that by adding AP we get AP = AC/(N+1). This suggests an iterative algorithm: for N = 2, there is a known construction of the midpoint B of AC by joining the points of intersection of two circles of equal radii centered at A and C. To trisect AC, reduce the radius of the circle at A to AB, which gives a new triangle ACD. The angle bisector of D will now divide AC in the ratio 1:2, such that AP = AC/3. Reduce the radius of the circle at A to AP and consider a new triangle ACD. The angles bisector will now divided AC in the ratio 1:3, and so on.
ACD, let DP be the angle bisector of angle D. Then
